As [tex]{x \to \ - 2^{-}, f(x) \to\ - \infty[/tex] and as [tex]{x \to \ - 2^{+}, f(x) \to\ \infty[/tex]
We have the function [tex]f(x) =\frac{x+8}{x^2+10x+16} \\[/tex]
- Vertical asymptotes are vertical lines that correspond to the zeroes of the denominator in a function.
Now for vertical asymptote put the denominator equals to zero
[tex]x^2+10x+16=0\\x^2+8x+2x+16\\x(x+8)+2(x+8)=0\\(x+2)(x+8)=0\\x=-2,-8[/tex]
So the vertical asymptote is at x=-2,-8
Now,
[tex]\lim_{x \to \-2^{-} } \frac{x+8}{x^2+10x+16} =- \infty \\ \lim_{x \to \-2^{+} } \frac{x+8}{x^2+10x+16} = \infty[/tex]
Therefore the correct option is second one that is
As [tex]{x \to \ - 2^{-}, f(x) \to\ - \infty[/tex] and as [tex]{x \to \ - 2^{+}, f(x) \to\ \infty[/tex]
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